Proof about components of a zero quintright.
Home.

The aim is to prove that if A = 0, then Aq = Ar = As = At = 0. By hypothesis:

0 = 2 Aq cos (18°) + 2 Ar cos (36°) + 2 As cos (54°) + 2 At cos (72°)

Substitute radical forms for the cosines and multiply both sides by two:

0 = Aq √(10 + √20) + Ar (√5 + 1) + As √(10 − √20) + At (√5 − 1)

Elimination of radicals. Rearrange the previous equation:

Ar (√5 + 1) − At (√5 − 1) = Aq √(10 + √20) + As √(10 − √20)

At the risk of introducing spurious roots, eliminate some radicals by squaring both sides:

Ar2 (6 + 2 √5) + At2 (6 − 2 √5) + 8 Ar At = Aq2 (10 + √20) + As2 (10 − √20) + 2 Aq As √80

6 Ar2 + 2 √5 Ar2 + 6 At2 − 2 √5 At2 + 8 Ar At = 10 Aq2 + 2 √5 Aq2 + 10 As2 − 2 √5 As2 + 8 √5 Aq As

Move all radicals to the left-hand member, and divide by two:

√5 (Ar2At2Aq2 + As2 − 4 Aq As) = − 3 Ar2 − 4 Ar At + 5 Aq2 − 3 At2 + 5 As2

Because √5 is irrational, but all of Aq, Ar, As, At are integral, the only hope of solution is when both of these equations are satisfied:

Ar2At2Aq2 + As2 − 4 Aq As = 0
− 3 Ar2 − 4 Ar At + 5 Aq2 − 3 At2 + 5 As2 = 0

Equate the two left-hand members and simplify:

Ar2At2Aq2 + As2 − 4 Aq As = − 3 Ar2 − 4 Ar At + 5 Aq2 − 3 At2 + 5 As2
2 Ar2 + At2 + 2 Ar At = 3 Aq2 + 2 As2 + 2 Aq As

Modulo two. Filter this last equation into modulo two. First, discard any term that is patently even. Then recall that the product of odd numbers is odd, and the product of anything else is even:

At2 = 3 Aq2 [mod 2]
At = Aq [mod 2]

Now recall this equation:

Ar2At2Aq2 + As2 − 4 Aq As = 0

which in modulo two becomes:

Ar2 + At2 + Aq2 + As2 = 0 [mod 2]
Ar + At + Aq + As = 0 [mod 2]

At = Aq [mod 2] implies At + Aq = 0 [mod 2]. Substitute the latter to obtain Ar + As = 0 [mod 2] which implies Ar = As [mod 2]. Now there are four cases:

• Aq, Ar, As, At are all even.
• Aq and At are both even, but Ar and As are both odd.
• Aq and At are both odd, but Ar and As are both even.
• Aq, Ar, As, At are all odd.

Modulo four. Again recall this equation:

Ar2At2Aq2 + As2 − 4 Aq As = 0

which in modulo four becomes:

Ar2At2Aq2 + As2 = 0 [mod 4]

Consider the case where Aq and At are both even, but Ar and As are both odd. Recall that the square of an even number is a multiple of four, and the square of an odd number is one more than a multiple of four. The previous equation then becomes:

1 − 0 − 0 + 1 = 0 [mod 4]
2 = 0 [mod 4] -- FALSE

Similarly dismiss the opposite case, where Aq and At are both odd and Ar and As are both even. Now two cases are left:

• Aq, Ar, As, At are all even.
• Aq, Ar, As, At are all odd.

Modulo eight. Note that the square of an odd number is always one more than a multiple of eight. To see this, write the odd number as 2n + 1; its square S becomes 4n2 + 4n + 1.

• If n is even, write n = 2m, then S = 16m2 + 8m + 1 = 8(2m2 + m) + 1.
• If n is odd, write n = 2m + 1, then S = 4(2m + 1)2 + 4(2m + 1) + 1 = 16m2 + 24m + 9 = 8(2m2 + 3m + 1) + 1

Now consider what happens when all of Aq, Ar, As, At are odd. Recall, yet again, the following equation:

Ar2At2Aq2 + As2 − 4 Aq As = 0

which appears the same in modulo eight:

Ar2At2Aq2 + As2 − 4 Aq As = 0 [mod 8]

From the comments above pertaining to the square of an odd number, substitute 1 for each of the squares:

1 − 1 − 1 + 1 − 4 Aq As = 0 [mod 8]
− 4 Aq As = 0 [mod 8]

Since Aq and As are odd, write Aq = 2Bq + 1 and As = 2Bs + 1, where Bq and Bs are integers. Substitute:

− 4 (2Bq + 1) (2Bs + 1) = 0 [mod 8]
16BqBs + 8Bq + 8Bs − 4 = 0 [mod 8]

Discarding the obvious multiples of eight leaves:

−4 = 0 [mod 8] -- FALSE

Now one case is left:

• Aq, Ar, As, At are all even.

Reduction. Since all the As are even, then must exist integer Bs satisfying the equations Aq = 2Bq, Ar = 2Br, As = 2Bs and At = 2Bt. From the original hypothesis A = 0, hence 2B = 0, and after scalar division by two, B = 0. This means that B must be a solution in its own right, and all its components are even.

As the existence of solution A necessitates the existence of solution B = A ÷ 2, the existence of solution B implies the existence of a solution C = B ÷ 2 = A ÷ 4. From this result, one must conclude that all the components of A were multiples of four. By the same token, there must also be solution D = C ÷ 2 = B ÷ 4 = A ÷ 8. Now it seems that all the components of A were multiples of eight. Further, solution E = D ÷ 2, which surely exists, calls for the components of A to be multiples of sixteen.

This solution-generating process can continue indefinitely, meaning that each component of A must have infinitely many factors of two, which cannot happen for a finite nonzero integer. Hence, all the components of a solution to A = 0 are themselves zero.